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1.12: Brønsted-Lowry Acids and Bases (Review)

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    learning objective

    • recognize acids and bases

    The Brønsted-Lowry Theory of Acids and Bases

    In 1923, Danish chemist Johannes Brønsted and English chemist Thomas Lowry independently proposed new definitions for acids and bases, ones that focus on proton transfer. A Brønsted-Lowry acid is any species that can donate a proton (H+) to another molecule. A Brønsted-Lowry base is any species that can accept a proton from another molecule. In short, a Brønsted-Lowry acid is a proton donor (PD), while a Brønsted-Lowry base is a proton acceptor (PA).

    A Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor.

    Let us use the reaction of ammonia in water to demonstrate the Brønsted-Lowry definitions of an acid and a base. Ammonia and water molecules are reactants, while the ammonium ion and the hydroxide ion are products:

    \[\ce{NH3(aq) + H2O (ℓ) <=> NH^{+}4(aq) + OH^{−}(aq) }\label{Eq1}\]

    What has happened in this reaction is that the original water molecule has donated a hydrogen ion to the original ammonia molecule, which in turn has accepted the hydrogen ion. We can illustrate this as follows:

    alt

    Because the water molecule donates a hydrogen ion to the ammonia, it is the Brønsted-Lowry acid, while the ammonia molecule—which accepts the hydrogen ion—is the Brønsted-Lowry base. Thus, ammonia acts as a base in both the Arrhenius sense and the Brønsted-Lowry sense.

    Is an Arrhenius acid like hydrochloric acid still an acid in the Brønsted-Lowry sense? Yes, but it requires us to understand what really happens when HCl is dissolved in water. Recall that the hydrogen atom is a single proton surrounded by a single electron. To make the hydrogen ion, we remove the electron, leaving a bare proton. Do we really have bare protons floating around in aqueous solution? No, we do not. What really happens is that the H+ ion attaches itself to H2O to make H3O+, which is called the hydronium ion. For most purposes, H+ and H3O+ represent the same species, but writing H3O+ instead of H+ shows that we understand that there are no bare protons floating around in solution. Rather, these protons are actually attached to solvent molecules.

    The Hydronium IOn

    A proton in aqueous solution may be surrounded by more than one water molecule, leading to formulas like \(\ce{H5O2^{+}}\) or \(\ce{H9O4^{+}}\) rather than \(\ce{H3O^{+}}\). It is simpler, however, to use \(\ce{H3O^{+}}\) to represent the hydronium ion.

    alt

    With this in mind, how do we define HCl as an acid in the Brønsted-Lowry sense? Consider what happens when HCl is dissolved in H2O:

    \[\ce{HCl(g) + H_2O (ℓ) \rightarrow H_3O^{+}(aq) + Cl^{−}(aq) }\label{Eq2}\]

    We can depict this process using Lewis electron dot diagrams:

    alt

    Now we see that a hydrogen ion is transferred from the HCl molecule to the H2O molecule to make chloride ions and hydronium ions. As the hydrogen ion donor, HCl acts as a Brønsted-Lowry acid; as a hydrogen ion acceptor, H2O is a Brønsted-Lowry base. So HCl is an acid not just in the Arrhenius sense but also in the Brønsted-Lowry sense. Moreover, by the Brønsted-Lowry definitions, H2O is a base in the formation of aqueous HCl. So the Brønsted-Lowry definitions of an acid and a base classify the dissolving of HCl in water as a reaction between an acid and a base—although the Arrhenius definition would not have labeled H2O a base in this circumstance.

    • A Brønsted-Lowry acid is a proton (hydrogen ion) donor.
    • A Brønsted-Lowry base is a proton (hydrogen ion) acceptor.
    • All Arrhenius acids and bases are Brønsted-Lowry acids and bases as well. However, not all Brønsted-Lowry acids and bases are Arrhenius acids and bases.

    Example \(\PageIndex{1}\)

    Aniline (C6H5NH2) is slightly soluble in water. It has a nitrogen atom that can accept a hydrogen ion from a water molecule just like the nitrogen atom in ammonia does. Write the chemical equation for this reaction and identify the Brønsted-Lowry acid and base.

    Solution

    C6H5NH2 and H2O are the reactants. When C6H5NH2 accepts a proton from H2O, it gains an extra H and a positive charge and leaves an OH− ion behind. The reaction is as follows:

    \[\ce{C6H5NH2(aq) + H2O(ℓ) <=> C6H5NH3^{+}(aq) + OH^{−}(aq)} \nonumber\]

    Because C6H5NH2 accepts a proton, it is the Brønsted-Lowry base. The H2O molecule, because it donates a proton, is the Brønsted-Lowry acid.

    Exercise \(\PageIndex{1}\)

    Identify the Brønsted-Lowry acid and the Brønsted-Lowry base in this chemical equation.

    \[\ce{H2PO4^{-} + H_2O <=> HPO4^{2-} + H3O^{+}}\]

    Answer:
    Brønsted-Lowry acid: H2PO4-; Brønsted-Lowry base: H2O

    Exercise \(\PageIndex{2}\)

    Which of the following compounds is a Bronsted-Lowry base?

    1. HCl
    2. HPO42-
    3. H3PO4
    4. NH4+
    5. CH3NH3+
    Answer:

    A Brønsted-Lowry Base is a proton acceptor, which means it will take in an H+. This eliminates \(\ce{HCl}\), \(\ce{H3PO4}\) , \(\ce{NH4^{+}}\) and \(\ce{CH_3NH_3^{+}}\) because they are Bronsted-Lowry acids. They all give away protons. In the case of \(\ce{HPO4^{2-}}\), consider the following equation:

    \[\ce{HPO4^{2-} (aq) + H2O (l) \rightarrow PO4^{3-} (aq) + H3O^{+}(aq) } \nonumber\]

    Here, it is clear that HPO42- is the acid since it donates a proton to water to make H3O+ and PO43-. Now consider the following equation:

    \[ \ce{ HPO4^{2-}(aq) + H2O(l) \rightarrow H2PO4^{-} + OH^{-}(aq)} \nonumber\]

    In this case, HPO42- is the base since it accepts a proton from water to form H2PO4- and OH-. Thus, HPO42- is an acid and base together, making it amphoteric.

    Since HPO42- is the only compound from the options that can act as a base, the answer is (b) HPO42-.

    Conjugate Acid-Base Pair

    In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), yielding the conjugate base of water, \(\ce{OH^-}\), and the conjugate acid of ammonia, \(\ce{NH4+}\):

    This figure has three parts in two rows. In the first row, two diagrams of acid-base pairs are shown. On the left, a space filling model of H subscript 2 O is shown with a red O atom at the center and two smaller white H atoms attached in a bent shape. Above this model is the label “H subscript 2 O (acid)” in purple. An arrow points right, which is labeled “Remove H superscript plus.” To the right is another space filling model with a single red O atom to which a single smaller white H atom is attached. The label in purple above this model reads, “O H superscript negative (conjugate base).” Above both of these red and white models is an upward pointing bracket that is labeled “Conjugate acid-base pair.” To the right is a space filling model with a central blue N atom to which three smaller white H atoms are attached in a triangular pyramid arrangement. A label in green above reads “N H subscript 3 (base).” An arrow labeled “Add H superscript plus” points right. To the right of the arrow is another space filling model with a blue central N atom and four smaller white H atoms in a tetrahedral arrangement. The green label above reads “N H subscript 3 superscript plus (conjugate acid).” Above both of these blue and white models is an upward pointing bracket that is labeled “Conjugate acid-base pair.” The second row of the figure shows the chemical reaction, H subscript 2 O ( l ) is shown in purple, and is labeled below in purple as “acid,” plus N H subscript 3 (a q) in green, labeled below in green as “base,” followed by a double sided arrow arrow and O H superscript negative (a q) in purple, labeled in purple as “conjugate base,” plus N H subscript 4 superscript plus (a q)” in green, which is labeled in green as “conjugate acid.” The acid on the left side of the equation is connected to the conjugate base on the right with a purple line. Similarly, the base on the left is connected to the conjugate acid on the right side.

    In the reaction of ammonia with water to give ammonium ions and hydroxide ions, ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are \(NH_4^+/NH_3\) and \(H_2O/OH^−\).

    alt
    Figure \(\PageIndex{1}\). The strongest acids are at the bottom left, and the strongest bases are at the top right. The conjugate base of a strong acid is a very weak base, and, conversely, the conjugate acid of a strong base is a very weak acid.
    alt
    Figure \(\PageIndex{1}\): The Relative Strengths of Some Common Conjugate Acid–Base Pairs

    The strongest acids are at the bottom left, and the strongest bases are at the top right. The conjugate base of a strong acid is a very weak base, and, conversely, the conjugate acid of a strong base is a very weak acid.

    Example \(\PageIndex{2}\)

    Identify the conjugate acid-base pairs in this equilibrium.

    \[\ce{CH3CO2H + H2O <=> H3O^{+} + CH3CO2^{-}} \nonumber\]

    Solution

    Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, \(H_3O^+\) is the acid that donates a proton to the acetate ion, which acts as the base.

    Once again, we have two conjugate acid–base pairs:

    • the parent acid and its conjugate base (\(CH_3CO_2H/CH_3CO_2^−\)) and
    • the parent base and its conjugate acid (\(H_3O^+/H_2O\)).

    alt

    Example \(\PageIndex{3}\)

    Identify the conjugate acid-base pairs in this equilibrium.

    \[(CH_{3})_{3}N + H_{2}O\rightleftharpoons (CH_{3})_{3}NH^{+} + OH^{-} \nonumber\]

    Solution

    One pair is H2O and OH−, where H2O has one more H+ and is the conjugate acid, while OH− has one less H+ and is the conjugate base.

    The other pair consists of (CH3)3N and (CH3)3NH+, where (CH3)3NH+ is the conjugate acid (it has an additional proton) and (CH3)3N is the conjugate base.

    Exercise \(\PageIndex{3}\)

    Identify the conjugate acid-base pairs in this equilibrium.

    \[\ce{NH2^{-} + H2O\rightleftharpoons NH3 + OH^{-}} \nonumber\]

    Answer:
    H2O (acid) and OH− (base); NH2− (base) and NH3 (acid)

    Contributors and Attributions

    1.12: Brønsted-Lowry Acids and Bases (Review) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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